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Lyrx [107]
3 years ago
9

What are noble gases ? Explain each example.​

Physics
1 answer:
victus00 [196]3 years ago
6 0

Answer:

18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og)

Explanation:

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40 POINTS AND RLLY EASY
Fynjy0 [20]

A. Jupiter largest and most massive planet in the solar system

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Jupiter has at least 61 moons orbiting it, some of which are very large

Jupiter located closer to the sun than Kuiper Belt

<h3>Which option that best describes the planet Jupiter?</h3>

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8 0
2 years ago
A 0.200-kg cube of ice (frozen water) is floating in glycerine.The gylcerine is in a tall cylinder that has inside radius 3.90 c
natita [175]

Answer:

Part a)

h = 0.86 cm

Part b)

Level will increase

Explanation:

Part a)

Mass of the ice cube is 0.200 kg

Now from the buoyancy force formula we know that weight of the ice is counter balanced by buoyancy force on the ice

So here we will have

mg = \rho V_{displaced} g

V_{displaced} = \frac{m}{\rho}

V_{displaced} = \frac{0.200}{1260} = 1.59 \times 10^{-4} m^3

now as we know that ice will melt into water

so here volume of water that will convert due to melting of ice is given as

V\rho_w = m_{ice}

V = \frac{0.200}{1000} = 2\times 10^{-4} m^3

So here extra volume that rise in the level will be given as

\Dleta V = V - V_{displaced}

\pi r^2 h = 2\times 10^{-4} - 1.59 \times 10^{-4}

(\pi (0.039^2) h = 0.41 \times 10^{-4}

h = 0.86 cm

Part b)

Since volume of water that formed here is more than the volume that is displaced by the ice so we can say that level of liquid in the cylinder will increase due to melting of ice

5 0
3 years ago
In a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section and a
timofeeve [1]

Answer:

a) F₁ = 1.48 x 10³ N

b) P = 1.88*10⁵ Pa

c) The work  is equal in both pistons

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F₁) on a small area piston (A₁), then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F₂) can be exerted that is proportional to the area(A₂) of the piston.

Pressure is defined as the force per unit area:

P=\frac{F}{A}  Formula (1)

P₁=P₂

\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} } Formula (2)

Data

r₁= 5 cm = 0.05 m

r₂= 15 cm = 0.15 m

F₂=  13300N

Area of the pistons (A₁,A₂)

A=π*r² : Area of the circle

A₁ = π*(0.05)²=7.85*10⁻³ m²

A₂= π*(0.15)²= 70.69*10⁻³ m²

a) Force that compressed air must exert to lift a car weighing 13300 N

We replace data in the formula (2)

\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} }

F_{1} = \frac{13300*7.85*10^{-3} }{70.69*10^{-3} }

F₁ =  1.48 x 10³ N

b) Air pressure produced by F₁

We replace data in the formula (1)

P=\frac{F}{A}

F₁ =  1.48 x 10³ N , A₁ = 7.85*10⁻³ m²

P=\frac{1.48*10^{3} }{7.85*10^{-3} }

P= 1.88*10⁵ Pa

c)The volume of liquid displaced by the small piston is distributed in a thin layer on the large piston, so that the product of the force by the displacement (the work) is equal in both pistons.

3 0
3 years ago
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